Let the line L pass through the point (-3, 5, 2) and make equal angles with the positive c…

Question

Let the line L pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :

Accepted Answer

Equation line is : $\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda $

herefore  General point R on line is $r(\lambda -3,\lambda +5,\lambda +2)$

$\begin{gathered} \overset{	o \leftarrow }{PR}=(\lambda -1,\lambda +5-r,\lambda +1) \ \overset{	o \leftarrow }{PR}=\vec{d}=0 \ \Rightarrow (\lambda -1)1+(\lambda +5-r)1+(\lambda +1)1=0 \end{gathered}$

$\begin{gathered} \Rightarrow 3\lambda +5=0 \ \Rightarrow \lambda =\frac{r-5}{3} \ 	herefore R\equiv (\frac{r-5}{3}-3,\frac{r-5}{3}+5,\frac{r-5}{3}+2) \ 	herefore R\equiv (\frac{r-14}{3},\frac{r+10}{3},\frac{r+1}{3}) \ PR=\sqrt{\frac{14}{3}}\Rightarrow (PR{)}^{2}=\frac{14}{3} \ \Rightarrow (\frac{r-14}{3}+2)+{(\frac{r+10}{3}-r)}^{2}+{(\frac{r+1}{3}-1)}^{2}=\frac{14}{3} \ \Rightarrow \frac{(r-8{)}^{2}}{9}+\frac{(10-2r{)}^{2}}{9}+\frac{(r-2{)}^{2}}{9}=\frac{14}{3} \ \Rightarrow ({r}^{2}-16r+64)+(100+4{r}^{2}-40r)+({r}^{2}-4r+4)=42 \ \Rightarrow 6{r}^{2}-60r+126=0 \ \Rightarrow {r}^{2}-10r+21=0 \ \Rightarrow r=3,7 \ r=10 \end{gathered}$

Solutions · 1